0

Let $p\geq 2$ be a natural number. Define $\equiv_p$ on $\mathbb{Z}$ by declaring $x\equiv_p$ if and only if there exists $k\in\mathbb{Z}$ such that $x-y=pk.$ For equivalence classes $[a]$ and $[b]$ define their product to be: $$[a].[b]=[a.b]$$ Prove that this is well-defined. That is, show that if $[a]=[a']$ and $[b]=[b']$, then $[a].[b]=[a'].[b']$

So I know that $a-a'=pn$ and $b-b'=pm$ and that I think I am trying to show that $[a.b]=[a'.b']$, but I am not sure how to reach that. I would appreciate the help.

KelKel23
  • 169
  • 2
    $$ab-a'b' = a(b-b') + b'(a-a') = apm + bpn = p(am+bn)$$ – Crostul May 03 '17 at 21:02
  • 1
    Aren't you using what you are trying to show? Can you please explain this a little more. I am confused as to what is being shown and what is assumed. – KelKel23 May 03 '17 at 21:10
  • I used what you wrote "So I know that...". That is the hypothesis. Now, there is that equality I wrote in the comment. That equality shows that $ab-a'b'$ is a multiple of $p$. What can you conclude? – Crostul May 03 '17 at 21:12
  • I am not sure how you got from $ab-a'b'$ to equal $a(b-b')+b'(a-a')$. I feel I am missing something super obvious here – KelKel23 May 03 '17 at 23:09
  • Expand it and you will see. – Crostul May 04 '17 at 06:04

0 Answers0